Solving Equations with Roots and Powers
Solving equations with roots and powers means undoing square roots, cube roots and powers to find the unknown, working in reverse BIDMAS order so you peel back one operation at a time. This page sets out the inverse-operation method step by step, with worked examples for square roots, squared terms, fractions and cube roots, and how to handle the plus-or-minus case when you undo a square. Below, free auto-marked practice questions for IGCSE Maths let you build speed across six rooms, from basic isolations to a mixed challenge.
How to Solve Equations with Roots and Powers (Reverse BIDMAS)
To get the unknown on its own, undo each operation in the reverse of BIDMAS order. Peel back the outside operations first and work inwards, doing the same step to both sides every time.
- Addition and subtraction first: undo a \(+k\) with \(-k\), and a \(-k\) with \(+k\).
- Then multiplication and division: undo \(\times k\) with \(\div k\), and \(\div k\) with \(\times k\).
- Then indices (powers and roots): square to undo a square root, cube to undo a cube root, and take a root to undo a power. When you undo a square, keep both answers \(x = \pm\sqrt{k}\) unless the context restricts \(x\).
- Brackets last: once the bracket or root is alone, deal with what is inside it.
- Always check your answer in the original equation: squaring can introduce solutions that do not fit.
Core ideas: inverse operations
Build it up, then unwind it
An equation is the unknown wrapped in layers. To solve \(\dfrac{\sqrt{x+2}}{2}=3\), start from the known value 3 and undo each layer in reverse, until \(x\) is on its own.
Worked examples
💡 Example 1: a root over a fraction
Solve \(\dfrac{\sqrt{x+2}}{2} = 3\).
\[\dfrac{\sqrt{x+2}}{2} = 3\]
Multiply both sides by 2: \(\ \sqrt{x+2} = 6\)
Square both sides: \(\ x+2 = 36\)
Subtract 2: \(x = 34\)
Check: \(\dfrac{\sqrt{34+2}}{2} = \dfrac{6}{2} = 3\) ✓
What's happening?
Undo in reverse BIDMAS: clear the \(\div 2\) first (so \(\times 2\)), then undo the root by squaring, then undo the \(+2\).
💡 Example 2: a multiple of a square root
Solve \(3\sqrt{x} = 21\).
\[3\sqrt{x} = 21\]
Divide both sides by 3: \(\ \sqrt{x} = 7\)
Square both sides: \(x = 49\)
Check: \(3\sqrt{49} = 3 \times 7 = 21\) ✓
What's happening?
The 3 multiplies the root, so divide first to leave the root alone, then square to undo it.
💡 Example 3: a squared term (the \(\pm\) case)
Solve \(5x^2 + 11 = 136\).
\[5x^2 + 11 = 136\]
Subtract 11: \(\ 5x^2 = 125\)
Divide by 5: \(\ x^2 = 25\)
Square root both sides: \(x = \pm 5\)
Check: \(5 \times 25 + 11 = 136\) ✓
What's happening?
Isolate \(x^2\) first (subtract 11, then divide by 5). When you square root, keep both signs: \((-5)^2 = 25\) too, so \(x = -5\) also works.
💡 Example 4: a cube root
Solve \(\sqrt[3]{3x - 2} = 4\).
\[\sqrt[3]{3x - 2} = 4\]
Cube both sides: \(\ 3x - 2 = 64\)
Add 2: \(\ 3x = 66\)
Divide by 3: \(x = 22\)
Check: \(\sqrt[3]{3 \times 22 - 2} = \sqrt[3]{64} = 4\) ✓
What's happening?
A cube root is undone by cubing. No \(\pm\) is needed: a cube root has just one real value.
🔑 Key points
- Undo in reverse BIDMAS: add/subtract, then multiply/divide, then powers and roots, then brackets.
- A square root is undone by squaring; a power is undone by the matching root.
- Squaring gives two answers: \(x^2 = k \Rightarrow x = \pm\sqrt{k}\), unless the context restricts \(x\).
- A cube root has one real value, so no \(\pm\).
- Always check in the original equation, especially after squaring.
⚠ Common pitfalls
- Forgetting the \(\pm\) when square rooting (\(x^2 = 25\) gives \(x = \pm 5\), not just \(5\)).
- Undoing operations in the wrong order: clear what is outside the root or power first.
- Not checking for extraneous solutions after squaring both sides.